Question: Simplify the following expression: $y = \dfrac{-3x^2- 1x+14}{-3x - 7}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(14)} &=& -42 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-42$ and add them together. Remember, since $-42$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-7}$ and ${b}$ is ${6}$ $ \begin{eqnarray} {ab} &=& ({-7})({6}) &=& -42 \\ {a} + {b} &=& {-7} + {6} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 {-7}x) + ({6}x +{14}) $ Factor out the common factors: $ x(-3x - 7) - 2(-3x - 7)$ Now factor out $(-3x - 7)$ $ (-3x - 7)(x - 2)$ The original expression can therefore be written: $ \dfrac{(-3x - 7)(x - 2)}{-3x - 7}$ We are dividing by $-3x - 7$ , so $-3x - 7 \neq 0$ Therefore, $x \neq -\frac{7}{3}$ This leaves us with $x - 2; x \neq -\frac{7}{3}$.